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# One-way analysis of variance - Assignment Example

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One-Way Analysis of Variance Name: Course: Professor: Institution: City and State: Date: One-way analysis of variance Introduction There are numerous forms of variance analysis including one-way Analysis of Variance (ANOVA), and two-way ANOVA among many others…

## Extract of sample"One-way analysis of variance"

Download file to see previous pages Basically, the basis of one-way ANOVA is to partition the sum of squares within and between classes. This method enables effective comparison of different classes simultaneously assuming the data is normally distributed. One way ANOVA is determined in three critical steps starting with obtaining squares for all classes of data. The degree of freedom, which is the total number of independent data that is considered to estimate a parameter, is also determined. Estimating degrees of freedom later on becomes effective in analysing null hypothesis. According to null hypothesis, the mean of classes under consideration is taken to be the same meaning that the variation within and between classes is not significantly different if not identical. This paper applies one-way ANOVA to analyze data for three categories of doctors. To analyse the variance, one-way ANOVA helps to establish the mean of individual groups, known as the treatment mean. Further, the grand mean, which is the mean for the entire data, is also computed. A scatter diagram (data on appendix) No. of years in NHS only (x-axis) Perform a one-way analysis of variance, recording all your interim calculations. Treatment mean for the three groups is: NHS only-11.25, private practice only-25.33 and both NHS and private practice-21.92. Grand mean= (11.25+25.33+21.92)/3 = 19.5 Estimate the treatment effects of the three groups. =11.25-19.5=-8.25 =25.33-19.5=5.83 =21.92-19.5=2.42 The researcher should then compute one-way ANOVA to determine whether the differences in effects are significant. To determine the variance, the following formula is used: One-way ANOVA, MS Total = MS Total/ (J-1) = (SS Within +SS between)/ (N-1) MS within estimates variability within a group, it is also known as SS residue or SS error. N is Degree of Freedom (D.F) calculated as; N-1, where N is the total number of observation within individual group. MS within= SS within/ D.F (N-1) On the other hand, MS between estimates variability between the groups, it is also known as SS explained since it shows variability explained by group membership. J is Degrees of Freedom (D.F) calculated as; J-1, where J is the total number of observations in all groups. MS between= SS between/ D.F (J-1) Ti=135, Tii=304, Tiii=263 (i) (?y) ^2 =702^2 = 13,689 N 36 (ii) ?Y^2= 12^2++27^2+1^2....+37^2= 19,578 (iii) ?Ti^2 = 135^2+ 304^2+ 263^2 = 1,518.75 +7,701.33+5,764.08 = 14,984.16 N 12 12 12 SS Within= 19,578-14,984.16 = 4,593.84 SS Between=14,984.16- 13,689 =1,295.16 SS Total= 19,578- 13,689= 5,889 Therefore: MS Total= SS Total/ (N-1) =5,889/36 =163.58 MS Between= SS Between/ (J-1) =1,295.16/2= 647.58 MS Within= SS Within/ (N-1) =4,593.84/ (36-3) =139.2 Source SS D.F Mean Square F Treatment SS Between= 1,295.16 J-1=2 SS Between/(J-1) =647.58 = MS Between MS Within = 4.7 Error SS Within= 4,593.84 N-J=33 SS Within/(N-1) =139.2 Total SS Total= 5,889 N-1=35 SS Total/(N-1) =168.26 Step1: Ho= ?= ?= ?, that is, treatments are equally effective Step2: An F statistic is appropriate measure, since the dependent variable is continuous and there are more than one group. Step 3: Since ? = 0.05 and D.F= 2, 33, accept Ho if F2, 33 < 19.4 Step4: The computed value of F-statistic is 4.7 Step 5: Accept H0. The treatments are equally effective. Explain what your results mean in a way that a non-statistician could understand. As mentioned above, one-way ANOVA seeks to compare two or more classes of data in order to determine if ...Download file to see next pagesRead More
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